`2.68xx10^(-3)` moles of solution containing anion `A^(n+)` require `1.61xx10^(-3)` moles of `MnO_(4)^(-)` for oxidation of `A^(n+)` to `AO_(3)^(-)` in acidic medium. What is the value of `n`?

Explanation

To solve the problem, we need to determine the value of \( n \) in the oxidation reaction of the anion \( A^{n+} \) to \( AO_3^{-} \) when reacting with \( MnO_4^{-} \) in acidic medium. Here’s a step-by-step solution:

### Step 1: Identify the oxidation and reduction reactions
The oxidation half-reaction involves the conversion of \( A^{n+} \) to \( AO_3^{-} \). The reduction half-reaction involves the conversion of \( MnO_4^{-} \) to \( Mn^{2+} \) in acidic medium.

### Step 2: Determine the oxidation state changes
1. For \( MnO_4^{-} \):
- The oxidation state of Mn in \( MnO_4^{-} \) can be calculated as follows:
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